ash.edu/mod/quiz/attempt.php?attempt=9481336 Question 1 Not yet answered Marked

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ash.edu/mod/quiz/attempt.php?attempt=9481336 Question 1 Not yet answered Marked out of 2.00 A Beaker is divided into two chambers, A and B. by a semi-permeable membrane lying across the middle of the beaker Chamber A contains the following concentration of ions: (Na ) = 13 mM. (K') = 100 mM, (CI) = 110 mM Chamber B contains the following concentration of ions: (Na ) = 83 mM. (K) = 10 mM, (CI) = 110 mM If the membrane is permeable only to Na', what is the transmembrane potential difference of Chamber A with respect to Chamber B? Answer in whole numbers, with no decimal places. In your answer give only the numerical (asbolute) value - Le. ignore the sign of the answer. Answer: Question 2 Not yet answered Marked out of 2.00 A Benkor is divided into two chambers, A and B, by a semi-permeable membrane lying across the middle of the beaker Chamber A contains the following concentration of ions: {Na ) = 58 mM., (K') = 150 mM, (CI) = 100 mM, (A ) = 106 mM Chamber B contains the following concentration of ions: {Na ) = 200 mM.. (K) = 4 mM, (CI) = 50 mM, (A) = 152 MM If the membrane is permeable only to K , what is the transmembrane potential difference of Chamber A relative to Chamber B? Answer in whole numbers, with no decimal places. In your answer give only the numerical (asbolute) value - ie ignore the sign of the answer. Answer: al 1 .docx Accessibility Docum....doc - not final.pdf Show All X

Explanation / Answer

This can be solved by Nernst equation. So the transmembrane potential difference will be

Ek = RT/ZF×ln(I+out/ I+in)

where R (universal gas content) is 2 calmol-1k-1

T is temperature Kelvin=0C+273

Z=valence of ionic species

F (Faraday's constant)= 2.3×104 Calv-1mol-1

I = ion concentration where o denotes ion concentration outside the cell and i denotes ion concentratiin inside the cell

where RT/ZF =25 mV at 200 C and ln=2.3×log10

So RT/ZF×2.3=58mV in 200C.

Chamber A of the beaker is considered as inner cellular environment and B as extracellular. While calculating potential difference we will take only the permeable ion as other ion concentration will be fixed and they were not allowed to move to chamber B.

Now, for Question 1,

Transmembrane potential difference for sodium is

Ex=58×log(83/13) =46mV.

For Question 2,

Transmembrane potential difference for potassium is

Ex= 58×log(4÷150)=-91mV

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